The maximum value of the function f(x)=2x^3-1 | vedclass

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(B) Given the set $A = \{x \mid x^2+20 \leq 9x\}$.Solving the inequality $x^2-9x+20 \leq 0$:$(x-4)(x-5) \leq 0$,which implies $x \in [4, 5]$.Now,consi

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$7$

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(B) Given the set $A = \{x \mid x^2+20 \leq 9x\}$.Solving the inequality $x^2-9x+20 \leq 0$:$(x-4)(x-5) \leq 0$,which implies $x \in [4, 5]$.Now,consider the function $f(x) = 2x^3-15x^2+36x-48$.Find the derivative $f'(x) = 6x^2-30x+36 = 6(x^2-5x+6) = 6(x-2)(x-3)$.For $x \in [4, 5]$,$f'(x) > 0$ because both $(x-2)$ and $(x-3)$ are positive in this interval.Since $f'(x) > 0$ for all $x \in [4, 5]$,the function $f(x)$ is strictly increasing on the interval $[4, 5]$.Therefore,the maximum value occurs at the right endpoint $x = 5$.$f(5) = 2(5)^3 - 15(5)^2 + 36(5) - 48$$f(5) = 2(125) - 15(25) + 180 - 48$$f(5) = 250 - 375 + 180 - 48 = 7$.

The maximum value of the function $f(x)=2x^3-15x^2+36x-48$ on the set $A=\{x \mid x^2+20 \leq 9x\}$ is

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